C#怎么读取Xml的属性

Question

C#XML

Answer 1

string expect="14022305";
XmlDocument doc = new XmlDocument();
doc.Load("<?xml version="1.0" encoding="gb2312" ?><xml><row opentime="2014-02-23 11:40:00" opencode="08,07,02,05,10" expect="14022316" abbdate="20140223" endtime="2014-02-23 11:40:00" order="16"  /><row opentime="2014-02-23 11:30:00" opencode="04,05,03,02,10" expect="14022315" abbdate="20140223" endtime="2014-02-23 11:30:00" order="15"  /><row opentime="2014-02-23 11:20:00" opencode="01,04,10,03,06" expect="14022314" abbdate="20140223" endtime="2014-02-23 11:20:00" order="14"  /><row opentime="2014-02-23 11:10:00" opencode="05,08,11,10,02" expect="14022313" abbdate="20140223" endtime="2014-02-23 11:10:00" order="13"  /><row opentime="2014-02-23 11:00:00" opencode="08,01,05,06,07" expect="14022312" abbdate="20140223" endtime="2014-02-23 11:00:00" order="12"  /><row opentime="2014-02-23 10:50:00" opencode="11,03,09,08,10" expect="14022311" abbdate="20140223" endtime="2014-02-23 10:50:00" order="11"  /><row opentime="2014-02-23 10:40:00" opencode="07,10,02,08,06" expect="14022310" abbdate="20140223" endtime="2014-02-23 10:40:00" order="10"  /><row opentime="2014-02-23 10:30:00" opencode="02,03,10,05,08" expect="14022309" abbdate="20140223" endtime="2014-02-23 10:30:00" order="9"  /><row opentime="2014-02-23 10:20:00" opencode="08,02,07,01,05" expect="14022308" abbdate="20140223" endtime="2014-02-23 10:20:00" order="8"  /><row opentime="2014-02-23 10:10:00" opencode="06,11,08,05,04" expect="14022307" abbdate="20140223" endtime="2014-02-23 10:10:00" order="7"  /></xml>");
XmlNodeList xnl = doc.SelectNodes("//row[@expect='" + expect + "']");
foreach (XmlNode xni in xnl)
{
   string opencode = xni.Attributes["opencode"].Value;
   string opentime = xni.Attributes["opentime"].Value;
   string abbdate = xni.Attributes["abbdate"].Value;
   string endtime = xni.Attributes["endtime"].Value;
   string order = xni.Attributes[""].Value;
   
}